LeetCode_Populating Next Right Pointers in Each Node

6430 단어 LeetCode
Given a binary tree



    struct TreeLinkNode {

      TreeLinkNode *left;

      TreeLinkNode *right;

      TreeLinkNode *next;

    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.



Initially, all next pointers are set to NULL.



Note:



You may only use constant extra space.

You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,



         1

       /  \

      2    3

     / \  / \

    4  5  6  7

After calling your function, the tree should look like:



         1 -> NULL

       /  \

      2 -> 3 -> NULL

     / \  / \

    4->5->6->7 -> NULL


방법1:constant extra space
/**

 * Definition for binary tree with next pointer.

 * struct TreeLinkNode {

 *  int val;

 *  TreeLinkNode *left, *right, *next;

 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}

 * };

 */

class Solution {

public:

    void connect(TreeLinkNode *root) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

      if( !root  || ( !root->left && !root->right)) return ;

        

        root->left->next = root-> right;

        if(root->next)

            root->right->next = root->next->left;    

        connect(root->left);

        connect(root->right);

        

    }

};

방법2: 공간 사용상 요구에 부합되지 않는 것 같다
/**

 * Definition for binary tree with next pointer.

 * struct TreeLinkNode {

 *  int val;

 *  TreeLinkNode *left, *right, *next;

 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}

 * };

 */

class Solution {

public:

    void connect(TreeLinkNode *root) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

      if(NULL == root) return ;

      queue<TreeLinkNode *> myqueue, tpqueue;

      TreeLinkNode *p,*pre;

      myqueue.push(root);

      while(!myqueue.empty()){

        pre = myqueue.front();myqueue.pop();

        if(pre->left) tpqueue.push(pre->left);

        if(pre->right) tpqueue.push(pre->right);

        while(!myqueue.empty()){

            p = myqueue.front();myqueue.pop();

            if(p->left) tpqueue.push(p->left);

            if(p->right) tpqueue.push(p->right);

            pre->next = p;pre = p;

        }

        

        myqueue.swap(tpqueue);

      }

    }

};

방법3: 방법1의 비귀속 실현, 제목의 공간 요구에 부합
class Solution {

public:

    void connect(TreeLinkNode *root) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

      if(NULL == root) return;

      TreeLinkNode *head = root;

      TreeLinkNode *tp;

      while(head->left){

        tp = head->left;

        while(head){

           if(head->left)

                head->left->next = head ->right;

            if(head ->right && head->next)

                head->right->next = head->next->left;

            head = head ->next;    

        }

        head = tp;

      }

    }

};

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